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If t and c are two complex numbers such that \[|t| \ne |c|,|t| = 1{\text{ }}and{\text{ }}z = \dfrac{{at + b}}{{t - c}},z = x + iy.\] locus of z is(where a,b are complex number.
A) Line segment
B) Straight line
C) Circle
D) None of these
Answer
![Verified]()
Verified
Verified
Hint: Since $|t| = 1$ we will try to separate the factor of ‘t’ from the expression $z = \dfrac{{at + b}}{{t - c}}$.
After separating the term ‘t’ we will take the modulus function of both the side and will eliminate ‘t’ using$ |t| = 1$.
Complete step by step solution:
Given data:
$
a,\;b,\;c,\; \text{and}\:{\text{ t}}\:{\text{are }}\:{\text{complex }}\:{\text{numbers }} \\
|t| = 1 \\
|t| \ne |c| \\
z = \dfrac{{at + b}}{{t - c}},z = x + iy. \\
$
Now solving the given equation, i.e.
$
z = \dfrac{{at + b}}{{t - c}} \\
\Rightarrow z(t - c) = at + b \\
\Rightarrow zt - at = b + cz \\
\Rightarrow t(z - a) = b + cz \\
\Rightarrow t = \dfrac{{b + cz}}{{z - a}} \\
$
Now, applying the modulus function on both sides of the equation, we obtain
$
\left| t \right| = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\
\sin ce{\text{ }}\left| t \right| = 1 \\
1 = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\
$
${\text{Now have also condition on}}$ ${\text{'c'}}$that \[\left| {\text{c}} \right| \ne {\text{1}}\]
$
1 = \left| {\dfrac{{c\left( {\dfrac{b}{c} + z} \right)}}{{z - a}}} \right| \\
we{\text{ }}know{\text{ }}that,{\text{ }}\left| {ab} \right| = \left| a \right|\left| b \right| \\
\Rightarrow 1 = \left| {\dfrac{{\dfrac{b}{c} + z}}{{z - a}}} \right|\left| c \right| \\
\Rightarrow \dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right| \\
$
Since b, c are complex numbers, then $\dfrac{b}{c}$ be also a complex number let ‘p’
$\dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + p}}{{z - a}}} \right|$
We know that,
$\left| {\dfrac{{z - \alpha }}{{z - \beta }}} \right| = A$
When $A \ne 1$, then the locus of z is a circle.
Here also $\left| c \right| \ne 1$ i.e.,
$\left| {\dfrac{{z + p}}{{z - a}}} \right| \ne 1$
Therefore the locus of z is a circle.
(C) The circle is the correct option.
Note: It is well known that $\left| z \right| = a,$ ‘a’ being a constant is an equation of a circle.
From the equation
$z = \dfrac{{at + b}}{{t - c}}$
Applying the modulus function on both sides of the equation, we obtain
$\left| z \right| = \left| {\dfrac{{at + b}}{{t - c}}} \right|$
We can say that the right-hand side will come to be a constant as a, b, c and t are some fixed complex number.
We can write it as
$\left| z \right| = f$, where f is constant, comes out to be an equation of a circle.
After separating the term ‘t’ we will take the modulus function of both the side and will eliminate ‘t’ using$ |t| = 1$.
Complete step by step solution:
Given data:
$
a,\;b,\;c,\; \text{and}\:{\text{ t}}\:{\text{are }}\:{\text{complex }}\:{\text{numbers }} \\
|t| = 1 \\
|t| \ne |c| \\
z = \dfrac{{at + b}}{{t - c}},z = x + iy. \\
$
Now solving the given equation, i.e.
$
z = \dfrac{{at + b}}{{t - c}} \\
\Rightarrow z(t - c) = at + b \\
\Rightarrow zt - at = b + cz \\
\Rightarrow t(z - a) = b + cz \\
\Rightarrow t = \dfrac{{b + cz}}{{z - a}} \\
$
Now, applying the modulus function on both sides of the equation, we obtain
$
\left| t \right| = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\
\sin ce{\text{ }}\left| t \right| = 1 \\
1 = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\
$
${\text{Now have also condition on}}$ ${\text{'c'}}$that \[\left| {\text{c}} \right| \ne {\text{1}}\]
$
1 = \left| {\dfrac{{c\left( {\dfrac{b}{c} + z} \right)}}{{z - a}}} \right| \\
we{\text{ }}know{\text{ }}that,{\text{ }}\left| {ab} \right| = \left| a \right|\left| b \right| \\
\Rightarrow 1 = \left| {\dfrac{{\dfrac{b}{c} + z}}{{z - a}}} \right|\left| c \right| \\
\Rightarrow \dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right| \\
$
Since b, c are complex numbers, then $\dfrac{b}{c}$ be also a complex number let ‘p’
$\dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + p}}{{z - a}}} \right|$
We know that,
$\left| {\dfrac{{z - \alpha }}{{z - \beta }}} \right| = A$
When $A \ne 1$, then the locus of z is a circle.
Here also $\left| c \right| \ne 1$ i.e.,
$\left| {\dfrac{{z + p}}{{z - a}}} \right| \ne 1$
Therefore the locus of z is a circle.
(C) The circle is the correct option.
Note: It is well known that $\left| z \right| = a,$ ‘a’ being a constant is an equation of a circle.
From the equation
$z = \dfrac{{at + b}}{{t - c}}$
Applying the modulus function on both sides of the equation, we obtain
$\left| z \right| = \left| {\dfrac{{at + b}}{{t - c}}} \right|$
We can say that the right-hand side will come to be a constant as a, b, c and t are some fixed complex number.
We can write it as
$\left| z \right| = f$, where f is constant, comes out to be an equation of a circle.
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